Backscatter communication system efficiency with diffusing surfaces

Abstract

In an ambient backscatter communication system, the waves generated by a source are reflected by a tag, in a variable manner in time. Therefore, the tag can transmit a message to a reader, without generating any radio wave and without battery. As a consequence, such a communication system is a promising technology for ultra-low energy wireless communications. In the simplest implementation of such a system, the tag sends a binary message by oscillating between two states and the reader detects the bits by comparing the two distinct received powers. In this paper, for the first time, we propose to analyze the impact of the shape of diffusing flat panel surfaces that diffuse in all directions, on an ambient backscatter communication system. We establish the analytical closed form expression of the power contrast in the presence of flat panels, by considering a rectangular surface and a disk-shaped surface, and we show that diffusing surfaces improve the power contrast. Moreover, our approach allows us to express the contrast to noise ratio, and therefore to establish the BER performance. Furthermore, we show that it makes it possible to improve the energetic performance, thanks to diffusing surfaces. For any configuration characterized by a fixed source, tag and reader, we moreover determine the precise locations of diffusing surfaces, which induce a maximum efficiency of the surfaces, whatever the wavelength. Furthermore, we show that it becomes possible to easily determine an optimal frequency which maximizes the contrast power, thanks to the expression of the contrast power.

Appendices

Appendix A. System geometric expressions

We denote $\psi = (\widehat{\overrightarrow{r_{0}},\overrightarrow{z}})$, $\phi = (\widehat{\overrightarrow{r_{0}},\overrightarrow{x}})$, $\psi ^{\prime } = (\widehat{\overrightarrow{{r^{\prime }}_{0}},\overrightarrow{z}})$ and $\phi ^{\prime } = (\widehat{\overrightarrow{{r^{\prime }}_{0}},\overrightarrow{x}})$. Considering a vector $\overrightarrow{r_{j}}$ = $\overrightarrow{SM_{j}}$ where $M_{j}$ is located at any location of the surface $W_{1}$, we can write the expression

$$\begin{aligned} \overrightarrow{r_{0}} + \overrightarrow{z} + \overrightarrow{x} = \overrightarrow{r_{j}} \end{aligned}$$

(A1)

where $\overrightarrow{z}$ and $\overrightarrow{x}$ represent the projections of $\overrightarrow{r_{j}}$ on the axis z and x axis. Considering $x_{j}$ and $z_{j}$ the coordinates of $\overrightarrow{r_{j}}$ according the axes x and z, we have

$$\begin{aligned} {r_{j}^{2}}= & {} {r_{0}^{2}} + {z_{j}^{2}} + {x_{j}^{2}} + 2 r_{0} z_{j} \cos \psi + 2 r_{0} x_{j} \cos \phi \\{} & {} + 2 z_{j} x_{j} \cos (\widehat{\overrightarrow{z},\overrightarrow{x}})\\= & {} {r_{0}^{2}} \left( 1+ 2 \frac{z_{j} \cos \psi }{r_{0}} + 2 \frac{x_{j} \cos \phi }{r_{0}} + \frac{{z_{j}^{2}} + {x_{j}^{2}} }{ {r_{0}^{2}}}\right) \end{aligned}$$

Therefore

$$\begin{aligned} r_{j}\approx & {} r_{0}+ 2 \frac{1}{2} z_{j} \cos \psi + 2 \frac{1}{2} x_{j} \cos \phi + \frac{{z_{j}^{2}}+ {x_{j}^{2}}}{2 r_{0}}\nonumber \\\approx & {} r_{0}+ z_{j} \cos \psi + x_{j} \cos \phi \end{aligned}$$

(A2)

due to $\cos (\widehat{\overrightarrow{z},\overrightarrow{x}}) = 0$ and considering that ${z_{j}^{2}} + {x_{j}^{2}}<< {r_{0}^{2}}$. We do the analog development considering the ray $r^{\prime }_{j}$ :

$$\begin{aligned} \overrightarrow{r^{\prime }_{j}} = \overrightarrow{- r^{\prime }_{0}} + \overrightarrow{z} + \overrightarrow{x} \end{aligned}$$

(A3)

So we have

$$\begin{aligned} {r^{\prime }}_{j}^{2}= & {} {r^{\prime }}_{0}^{2} + {z_{j}^{2}} + {x_{j}^{2}}\nonumber \\- & {} 2 {r^{\prime }}_{0} z_{j} \cos \psi ^{\prime } - 2 {r^{\prime }}_{0} x_{j} \cos \phi ^{\prime } + 2 z_{j} x_{j} \cos (\widehat{\overrightarrow{z},\overrightarrow{x}})\nonumber \\= & {} {r^{\prime }}_{0}^{2} \left( 1- 2 \frac{z_{j} \cos \psi ^{\prime }}{{r^{\prime }}_{0}} - 2 \frac{x_{j} \cos \phi ^{\prime }}{{r^{\prime }}_{0}} + \frac{{z_{j}^{2}} + {x_{j}^{2}} }{{r^{\prime }}_{0}^{2}}\right) \end{aligned}$$

(A4)

Therefore

$$\begin{aligned} {r^{\prime }}_{j}\approx & {} {r^{\prime }}_{0} - 2 \frac{1}{2} z_{j} \cos \psi ^{\prime } - 2 \frac{1}{2} x_{j} \cos \phi ^{\prime } + \frac{{z_{j}^{2}}+ {x_{j}^{2}}}{2 {r^{\prime }}_{0}}\nonumber \\\approx & {} {r^{\prime }}_{0} - z_{j} \cos \psi ^{\prime } - x_{j} \cos \phi ^{\prime } \end{aligned}$$

(A5)

since $\cos (\widehat{\overrightarrow{z},\overrightarrow{x}}) = 0$ and considering that ${z_{j}^{2}} + {x_{j}^{2}}<< {r^{\prime }}_{0}^{2}$. So we can write in the first order

$$\begin{aligned} r_{j} + {r^{\prime }}_{j}= & {} r_{0}\!+\! {r^{\prime }}_{0} \!+\! z_{j} \cos \psi \!+\! x_{j} \cos \phi - z_{j} \cos \psi ^{\prime } - x_{j} \cos \phi ^{\prime }\nonumber \\= & {} r_{0} + {r^{\prime }}_{0} + z_{j} (\cos \psi - \cos \psi ^{\prime }) + x_{j} (\cos \phi - \cos \phi ^{\prime })\nonumber \\= & {} r_{0} + {r^{\prime }}_{0} + z_{j} \gamma + x_{j} \zeta \end{aligned}$$

(A6)

From (A2) and (A5) $r_{0} {r^{\prime }}_{0} \approx r_{j} {r^{\prime }}_{j}, \forall$ j = 0...N. We have

$$\begin{aligned} Y_{dif}= & {} \sum _{j=1}^{N} \sqrt{\frac{K^{sar}P}{{r_{j}^{2}}}} \sqrt{\frac{1}{{r^{\prime }}_{j}^{2}}} e^{iwt}e^{-i k(r_{j}+{r^{\prime }}_{j})}\nonumber \\\approx & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt}e^{-i k(r_{0}+{r^{\prime }}_{0})} \sum _{j} e^{-i(k z_{j} \gamma + k x_{j} \zeta )} \end{aligned}$$

(A7)

Therefore

$$\begin{aligned} Y_{dif}\approx & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt}e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \int _{-L}^{L} \int _{-H}^{H} \exp -i(k \gamma z) \exp -i(k \zeta x) dzdx\nonumber \\\approx & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt}e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \int _{-L}^{L} \exp -i(k \gamma z) dz \int _{-H}^{H} \exp -i(k \zeta x) dx\nonumber \\\approx & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \frac{e^{-i(k\gamma L)}- e^{i(k \gamma L)}}{-ik \gamma } \frac{e^{-i(k\zeta H)}- e^{i(k \zeta H)}}{-ik \zeta }\nonumber \\\approx & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \frac{4}{k^{2} \gamma \zeta } \sin (k\gamma L) \sin (k\zeta H) \end{aligned}$$

(A8)

Therefore $Y_{dif}$ can be approximated by the following expression

$$\begin{aligned} Y_{dif} = \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})} F_{plan} \end{aligned}$$

(A9)

where

$$\begin{aligned} F_{plan} = \frac{4}{k^{2} \gamma \zeta } \sin (k\gamma L) \sin (k\zeta H) \end{aligned}$$

(A10)

If the diffusing surface is a disk of radius $R_{d}$, we can write

$$\begin{aligned} Y_{dif}\approx & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt}e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \int _{Surf} \exp -i k(\gamma z + \zeta x) dzdx \nonumber \\\approx & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt}e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \int _{0}^{R_{d}} \int _{0}^{2 \pi } \exp -i k( \Omega \rho \cos \theta ) \rho d \rho d \theta \end{aligned}$$

(A11)

Where $\overrightarrow{\Omega } (\zeta , 0, \gamma )$ and $\overrightarrow{\rho } (x, 0, z)$. So we can write $\gamma z + \zeta x$ as the scalar product of $\overrightarrow{\rho }$ and $\overrightarrow{\Omega }$. And we can express $\overrightarrow{\rho }.\overrightarrow{\Omega }$ = $\rho \Omega \cos \theta$ where $\rho$ and $\Omega$ are the module of $\overrightarrow{\rho }$ and $\overrightarrow{\Omega }$ and where $\theta = \widehat{(\overrightarrow{\rho },\overrightarrow{\Omega })}$. So we can express the approximation

$$\begin{aligned} Y_{dif}= & {} \sqrt{\frac{K^{sar}P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2}}} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \int _{0}^{R_{d}} \int _{0}^{2 \pi } \exp -i k( \Omega \rho \cos \theta ) \rho d \rho d \theta \nonumber \\= & {} A_{dif} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \int _{0}^{R_{d}} \int _{0}^{2 \pi } \exp -i k( \Omega \rho \cos \theta ) \rho d \rho d \theta \nonumber \\= & {} A_{dif} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})}\nonumber \\{} & {} \times \int _{0}^{R_{d}} \rho d \rho 2 \pi \times \frac{1}{2 \pi } \int _{0}^{2 \pi } \exp -i (k \Omega \rho \cos \theta ) d \theta \nonumber \\= & {} A_{dif} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})} \int _{0}^{R_{d}} 2 \pi \times J_{0}(k\Omega \rho ) \rho d \rho \nonumber \\= & {} A_{dif} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})} 2 \pi \int _{0}^{k\Omega R_{d}} \frac{x}{k\Omega } J_{0}(x) \frac{dx}{k\Omega } \end{aligned}$$

(A12)

Since $x J_{0}(x) = \frac{d}{dx}(x J_{1}(x))$, where $J_{0} (x)$ and $J_{1} (x)$ are the Bessel functions defined as $J_{0} (x) = \frac{1}{2 \pi } \int _{0}^{2 \pi } e^{-i (x \cos \theta )} d \theta$ and $J_{1} (x) = \frac{1}{2 \pi } \int _{0}^{2 \pi } e^{-i (x \cos \theta -\theta )} d \theta$. So we have

$$\begin{aligned} Y_{dif} = A_{dif} e^{iwt} e^{-i k(r_{0}+{r^{\prime }}_{0})} \frac{2 \pi R_{d}}{k\Omega } J_{1}(k\Omega R_{d}) \end{aligned}$$

(A13)

Therefore, in the case of a disk, $F_{plan}$ is replaced by $F_{disk}$ in (A9)

$$\begin{aligned} F_{disk} = \frac{2 \pi R_{d}}{k\Omega } J_{1}(k\Omega R_{d}) \end{aligned}$$

(A14)

Appendix B. Power received by the reader directly from the source

Considering P the transmit power of the source, $S_{R}$ the power density received by the reader coming from the source, $A_{R}$ the apparent area of the reader, $G^{s}$ the gain of the source, we can express

$$\begin{aligned} P_{R}= & {} S_{R} A_{R}\nonumber \\= & {} \frac{P}{4 \pi r^{2}} G^{s} A_{R} \end{aligned}$$

(B1)

Considering $G^{r}$ the gain of the reader, we have

$$\begin{aligned} G^{r} = A_{R} \frac{4 \pi }{\lambda ^{2}} \end{aligned}$$

(B2)

where $\lambda$ is the wavelength of the transmitted signal. Therefore

$$\begin{aligned} A_{R} = G^{r} \frac{\lambda ^{2}}{4 \pi } \end{aligned}$$

(B3)

As a consequence we can write

$$\begin{aligned} P_{R} = \frac{P}{r^{2}} G^{s} G^{r} \frac{ \lambda ^{2}}{(4 \pi )^{2}} \end{aligned}$$

(B4)

Therefore

$$\begin{aligned} P_{R} = \frac{K^{sr}}{r^{2} } P \end{aligned}$$

(B5)

where

$$\begin{aligned} K^{sr} = \frac{G^{s} G^{r}\lambda ^{2}}{(4 \pi )^{2}} \end{aligned}$$

(B6)

Appendix C. Power reflected by the tag or the surface, and received by the reader

1.1 1) Power reflected by the tag

Considering $P_{t}$ the power reflected by the tag coming from the source, $S_{t}$ the power density received by the tag coming from the source, $A_{t}$ the apparent area of the tag, we can express

$$\begin{aligned} P_{t}= & {} S_{t} A_{t} \nonumber \\= & {} \frac{P}{4 \pi {r_{t}^{2}}} G^{s} A_{t} \end{aligned}$$

(C1)

Considering $G^{t}$ the gain of the tag, we have

$$\begin{aligned} G^{t} = A_{t} \frac{4 \pi }{\lambda ^{2}} \end{aligned}$$

(C2)

where $\lambda$ is the wavelength of the transmitted signal. Therefore

$$\begin{aligned} A_{t} = G^{t} \frac{\lambda ^{2}}{4 \pi } \end{aligned}$$

(C3)

As a consequence we can write

$$\begin{aligned} P_{t} = \frac{P}{4 \pi {r_{t}^{2}}} G^{s} G^{t} \frac{ \lambda ^{2}}{4 \pi } \end{aligned}$$

(C4)

1.2 2) Power received by the reader from the tag

Considering ${P_{R}^{T}}$ the power received by the reader coming from the tag, ${S_{R}^{T}}$ the power density received by the reader coming from the tag, $A_{R}$ the apparent area of the reader, $G^{r}$ the gain of the reader we can express

$$\begin{aligned} {P_{R}^{T}}= & {} {S_{R}^{T}} A_{R}\nonumber \\= & {} \frac{P_{t}}{4 \pi {r^{\prime }}_{t}^{2}} G^{t} A_{R}\nonumber \\= & {} \frac{P_{t}}{4 \pi {r^{\prime }}_{t}^{2}} G^{t} G^{r} \frac{ \lambda ^{2}}{4 \pi }\nonumber \\= & {} \frac{\frac{P}{{r_{t}^{2}}} G^{s} G^{t} \frac{ \lambda ^{2}}{(4 \pi )^{2}} }{4 \pi {r^{\prime }}_{t}^{2}} G^{t} G^{r} \frac{ \lambda ^{2}}{4 \pi }\nonumber \\= & {} \frac{P}{{r_{t}^{2}} {r^{\prime }}_{t}^{2} } G^{s} (G^{t})^{2} G^{r} \frac{ \lambda ^{4}}{(4 \pi )^{4}} \end{aligned}$$

(C5)

since we have

$$\begin{aligned} G^{r} = A_{R} \frac{4 \pi }{\lambda ^{2}}. \end{aligned}$$

(C6)

and thus

$$\begin{aligned} A_{R} = G^{r} \frac{\lambda ^{2}}{4 \pi }. \end{aligned}$$

(C7)

Therefore

$$\begin{aligned} {P_{R}^{T}} = \frac{P}{{r_{t}^{2}} {r^{\prime }}_{t}^{2} } G^{s} (G^{t})^{2} G^{r} \frac{ \lambda ^{4}}{(4 \pi )^{4}} \end{aligned}$$

(C8)

Therefore

$$\begin{aligned} {P_{R}^{T}} = \frac{K^{str}}{{r_{t}^{2}} {r^{\prime }}_{t}^{2} } P \end{aligned}$$

(C9)

where

$$\begin{aligned} K^{str} = \frac{G^{s} (G^{t})^{2} G^{r}\lambda ^{4}}{(4 \pi )^{4}} \end{aligned}$$

(C10)

1.3 3) Power received by the reader from the surface

In an analog way, we can express the power ${P_{R}^{d}}$, received by the reader coming from the surface

$$\begin{aligned} {P_{R}^{d}} = \frac{P}{{r_{0}^{2}} {r^{\prime }}_{0}^{2} } G^{s} (G^{d})^{2} G^{r} \frac{ \lambda ^{4}}{(4 \pi )^{4}} \end{aligned}$$

(C11)

Therefore

$$\begin{aligned} {P_{R}^{d}} = \frac{K^{sar}}{{r_{0}^{2}} {r^{\prime }}_{0}^{2} } P \end{aligned}$$

(C12)

where

$$\begin{aligned} K^{sar} = \frac{G^{s} (G^{d})^{2} G^{r}\lambda ^{4}}{(4 \pi )^{4}} \end{aligned}$$

(C13)

Appendix D. Location of the diffusing surface

Let consider the coordinates of the source S(0, 0, 0) the reader $R(x_{R},y_{R},z_{R})$ and the center of the diffusing surface $O_{1}(x,y,z)$. We have $r_{0}(x, y, z)$ and $r^{\prime }_{0}(x^{\prime }, y^{\prime }, z^{\prime })$ where

$$\begin{aligned} \left\{ \begin{array}{lll} x^{\prime } = x-x_{R}\\ y^{\prime } = y-y_{R} r\\ z^{\prime } = z-z_{R}. \end{array}\right. \end{aligned}$$

(D1)

From

$$\begin{aligned} \left\{ \begin{array}{ll} \cos \psi =\cos \psi ^{\prime }\\ \cos \phi =\cos \phi ^{\prime } \end{array}\right. \end{aligned}$$

we can write

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{x}{r_{0}} = \frac{x-x_{R}}{{r^{\prime }}_{0}}\\ \frac{z}{r_{0}} = \frac{z-z_{R}}{{r^{\prime }}_{0}}. \end{array}\right. \end{aligned}$$

(D2)

Let notice that in our system, the coordinate z is not equal to zero. Otherwise, it would imply $z_{R}=0$ (which is not the case in our system (Fig. 3)). The coordinate x can be equal to zero. In this case, $x_{R}=0$. Therefore, from (D2)

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{x^{2}}{{r_{0}^{2}}} = \frac{x^{\prime 2}}{{r^{\prime }}_{0}^{2}}\\ \frac{z^{2}}{{r_{0}^{2}}} = \frac{(z-z_{R})^{2}}{{r^{\prime }}_{0}^{2}}. \end{array}\right. \end{aligned}$$

So we have

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{x}{z} = \frac{x-x_{R}}{z-z_{R}}\\ z^{2}((y-y_{R})^{2}+(x-x_{R})^{2}) = (z-z_{R})^{2}(y^{2}+x^{2}), \end{array}\right. \end{aligned}$$

which can be expressed

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{x}{z} = \frac{x-x_{R}}{z-z_{R}}\\ z^{2}((y-y_{R})^{2}+ (x\frac{z-z_{R}}{z})^{2}) = (z-z_{R})^{2}(y^{2}+x^{2}). \end{array}\right. \end{aligned}$$

Therefore

$$\begin{aligned} \left\{ \begin{array}{ll} \frac{x}{z} = \frac{x-x_{R}}{z-z_{R}}\\ y-y_{R} = \pm y\frac{z-z_{R}}{z}. \end{array}\right. \end{aligned}$$

(D3)

So we have two cases.

1.1 1) The first case is expressed

$$\begin{aligned} \left\{ \begin{array}{ll} z = \frac{z_{R}}{x_{R}}x\\ z = \frac{z_{R}}{y_{R}}y \end{array}\right. \end{aligned}$$

where $x_{R}$ and $y_{R}$ are not equal to 0. Therefore, denoting $a =\frac{x_{R}}{z_{R}}$ and $b =\frac{y_{R}}{z_{R}}$

$$\begin{aligned} \left\{ \begin{array}{ll} x = a z \\ y = b z. \end{array}\right. \end{aligned}$$

In this case, the set of locations of $O_{1}$ is a line, intersection of 2 plans. The coordinates of a point of this line are expressed as z(a, b, 1) where z is a real number.

1.2 2) The second case is expressed

$$\begin{aligned} \left\{ \begin{array}{ll} x = a z \\ y = \frac{y_{R}}{2z - z_{R}}z. \end{array}\right. \end{aligned}$$

In this case, the set of locations of $O_{1}$ is given by the intersection of the plan $z = \frac{1}{a}x$ and the hyperbole $z = \frac{z_{R}}{2y - y_{R}}y$. It can be expressed $O_{1}(az, \frac{y_{R} z}{2z - z_{R}}, z)$ where z is a real number (not equal to $\frac{z_{R}}{2}$). However if $z=\frac{z_{R}}{2}$ from (D3) we have $y_{R}=0$. If $x_{R}$ = 0, therefore x = 0, so we have

$$\begin{aligned} \left\{ \begin{array}{ll} x = 0 \\ z^{2} (y-y_{R})^{2} = (z-z_{R})^{2} y^{2}. \end{array}\right. \end{aligned}$$

This case is equivalent to the precedent case, with the plan given by $x= 0$.